Question

Increasing and decreasing functions Find the intervals on which \(f\) is increasing and the intervals on which it is decreasing. $$f(x)=-2 x^{4}+x^{2}+10$$

Short answer

Based on the given information about the function f(x), identify the intervals on which the function is increasing and decreasing. Answer: The function f(x) is increasing on the intervals (-1/2, 0) and (0, 1/2), and decreasing on the intervals (-∞, -1/2) and (1/2, ∞).

Step-by-step solution

Find the first derivative

To find the first derivative of the function, we will use the power rule. Let's find the derivative of f(x) with respect to x: $$f'(x)=-8x^3+2x$$

Find the critical points

Critical points are the points where the first derivative is equal to zero or it's undefined (the latter doesn't apply in this case). Let's find the critical points by setting f'(x) equal to zero and solving for x: $$-8x^3+2x = 0$$ Factor out a common factor of 2x: $$2x(-4x^2+1) = 0$$ Now, we have two factors, and set them equal to zero: $$2x=0 \Longrightarrow x=0$$ $$-4x^2 + 1 = 0 \Longrightarrow x^2 = \frac{1}{4} \Longrightarrow x = \pm\frac{1}{2}$$ So, we have three critical points: \(x = -\frac{1}{2}, x = 0, x = \frac{1}{2}\).

Determine intervals of increase and decrease

To determine the intervals of increase and decrease, we'll test the sign of the first derivative within each interval between the critical points: 1. Interval \((-\infty, -\frac{1}{2})\): Choose a test point, for example \(x=-1\). Plug into the first derivative: $$f'(-1) = -8(-1)^3 + 2(-1) = -10 < 0$$ Since f'(-1) < 0, the function is decreasing in this interval. 2. Interval \((-\frac{1}{2}, 0)\): Choose a test point, for example \(x=-\frac{1}{4}\). Plug into the first derivative: $$f'\left(-\frac{1}{4}\right) = -8\left(-\frac{1}{4}\right)^3+2\left(-\frac{1}{4}\right) = 1 > 0$$ Since f'(-1/4) > 0, the function is increasing in this interval. 3. Interval \((0, \frac{1}{2})\): Choose a test point, for example \(x=\frac{1}{4}\). Plug into the first derivative: $$f'\left(\frac{1}{4}\right) = -8\left(\frac{1}{4}\right)^3+2\left(\frac{1}{4}\right)= 1 > 0$$ Since f'(1/4) > 0, the function is increasing in this interval. 4. Interval\((\frac{1}{2},\infty)\): Choose a test point, for example \(x=1\). Plug into the first derivative: $$f'(1) = -8(1)^3 + 2(1) = -6 < 0$$ Since f'(1) < 0, the function is decreasing in this interval. In conclusion, the function is increasing on the intervals \((-\frac{1}{2},0)\) and \((0,\frac{1}{2})\) and decreasing on the intervals \((-\infty,-\frac{1}{2})\) and \((\frac{1}{2},\infty)\).

Key concepts

First Derivative

The first derivative of a function gives us vital information about the behavior of the function's graph. In essence, it tells us how the function is changing at any given point. To find the first derivative of a function like \( f(x) = -2x^4 + x^2 + 10 \), we use calculus rules such as the power rule.
The power rule states that if you have a term in the form of \( x^n \), the derivative is \( nx^{n-1} \). Applying this to our function, we find the first derivative:

• For \( -2x^4 \), the derivative is \( -8x^3 \).
• For \( x^2 \), the derivative is \( 2x \).
• The constant term 10 has a derivative of 0 as constants do not change.

Putting it all together, the first derivative of \( f(x) \) is \( f'(x) = -8x^3 + 2x \).
This derivative function, \( f'(x) \), will help us determine where the original function "f" increases or decreases.

Critical Points

Critical points are places on the graph where the function changes its direction from increasing to decreasing or vice versa. To find them, we solve \( f'(x) = 0 \), which means the slope of our tangent to the function becomes zero, indicating a possible max or min point.
Let's solve the equation derived from the derivative:
\(-8x^3 + 2x = 0\)

• First, factor out \(2x\): \(2x(-4x^2+1) = 0\).
• Setting each factor equal to zero gives us the critical points: \(2x = 0\) leads to \(x=0\) and \(-4x^2 + 1 = 0\) simplifies to \(x = \pm\frac{1}{2}\).

Thus, the critical points are \(x = -\frac{1}{2}, x = 0, \text{and } x = \frac{1}{2}\). These points partition the number line into intervals to be examined for increases or decreases.

Intervals of Increase and Decrease

Once we've pinpointed the critical points, we divide the entire real number line into intervals based on these points. These intervals can then be tested for increasing or decreasing behavior. Testing involves picking a sample point from each interval and substituting it into the first derivative \( f'(x) \).
Here’s how it works:

• Interval \((-\infty, -\frac{1}{2})\): Choose \(x=-1\), since \(f'(-1)=-10<0\), the function decreases here.
• Interval \((-\frac{1}{2}, 0)\): Choose \(x=-\frac{1}{4}\), since \(f'(-\frac{1}{4})=1>0\), the function increases in this interval.
• Interval \((0, \frac{1}{2})\): Choose \(x=\frac{1}{4}\), again \(f'(\frac{1}{4})=1>0\), the function increases here as well.
• Interval \((\frac{1}{2}, \infty)\): Choose \(x=1\), \(f'(1)=-6<0\), showing the function decreases in this final interval.

Hence, the original function is increasing on the intervals \((-\frac{1}{2},0)\) and \((0,\frac{1}{2})\). It is decreasing on \((-\infty, -\frac{1}{2})\) and \((\frac{1}{2}, \infty)\). These intervals neatly provide a map of where the function climbs and falls on its curve.